2nd PUC Maths Previous Year Question Paper June 2019

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Karnataka 2nd PUC Maths Previous Year Question Paper June 2019

Time: 3 Hrs 15 Min
Max. Marks: 100

Part – A

Answer ALL the following questions: (10 × 1 = 10)

Question 1.
Let * be the binary operation on N given.by a * b = LCM of a and b, ∀ a ? b ∈ N.
Answer:
Given a * b = LCM of a and b ∀ a, b ∈ N 5*7 = 35

Question 2.
Find the principal value of cot^-1\(\left(-\frac{1}{\sqrt{3}}\right)\)
Answer:
Cot^-1\(\left(-\frac{1}{\sqrt{3}}\right)\) = π – Cot^-1\(\left(-\frac{1}{\sqrt{3}}\right)\)
[∵ cot^-1(-x) = π – cot^-1(x)]
= π – \(\frac{\pi}{3}\)
∴ cot^-1 \(\left(\frac{-1}{\sqrt{3}}\right)=\frac{2 \pi}{3}\)

Question 3.
If A = [a_ij] where elements are given by a_ij = \(\frac { 1 }{ 2 }\)|-3i + j| construct 2 × 2 matrix.
Answer:
Given a_ij = \(\frac { 1 }{ 2 }\) |-3i + j|
a₁₁ = \(\frac { 1 }{ 2 }\)|-3(1) + 1|= \(\frac { 1 }{ 2 }\)|-2| = 1
a₁₂ = \(\frac { 1 }{ 2 }\) |-3(1) + 2| = \(\frac { 1 }{ 2 }\)|-l| = \(\frac { 1 }{ 2 }\)
a₂₁ = \(\frac { 1 }{ 2 }\) |-3(2) + 2| = \(\frac { 1 }{ 2 }\) |-5| = \(\frac { 5 }{ 2 }\)
a₂₂ = \(\frac { 1 }{ 2 }\) |-3(2) + 2| = \(\frac { 1 }{ 2 }\) |-4| = 2

Question 4.
Find a value of x for which

Answer:

3 – x² = 3 – 8 ⇒ x² = 8 ⇒ x = ±2\(\sqrt{2}\)

Question 5.
If y = cos^-1 (e^x), find \(\frac{d y}{d x}\).
Answer:
y = cos^-1 (e^x) (diff wrt x)

Question 6.
Find \(\int \sec ^{2}(7-4 x) d x\)
Answer:

Question 7.
If â = \(\frac{1}{\sqrt{14}}\) (2î + 3ĵ + k̂), then write the directions cosines of â.
Answer:
â = \(\frac{1}{\sqrt{14}}\) (2î + 3ĵ + k̂)
∴ Directions cosines are :
\(\frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{1}{\sqrt{14}}\)

Question 8.
Find the intercepts cut off by the plane 2x + y – z = 5
Answer:
2x + y – z = 5 …..(1) ÷ 5
\(\frac{2 x}{5}+\frac{y}{5}-\frac{z}{5}=1\)
or \(\frac{x}{(5 / 2)}+\frac{y}{5}+\frac{z}{(-5)}=1\)
∴ x – intercept = \(\frac { 5 }{ 2 }\) ; y – intercept = 5, z – intercept = -5

Question 9.
Define feasible region in a linear programming problem.
Answer:
Feasible region is defined as the common region determined by all the constraints including non – negative constraints x ≥ 0, y ≥ 0 of a linear programming problem is called as Feasible region.

Question 10.
If P(A) = \(\frac { 3 }{ 5 }\) and P(B) = \(\frac { 1 }{ 5 }\) find P(A ∩ B), where A and B are independent events.
Answer:
P(A) = \(\frac { 3 }{ 5 }\); P(B) = \(\frac { 1 }{ 5 }\)
P(A ∩ B) = P(A).P(B)
[Since A and B are independent events]
= \(\frac{3}{5} \cdot \frac{1}{5}\) = \(\frac { 3 }{ 25 }\)

Part-B

Answer any TEN questions: (10 x 2 = 20)

Question 11.
Find g.f and f.g, if (x) = 8x³ and g(x) = x^1/3
Answer:
Given
f(x) = 8x³ ,g(x)= x^1/3
fog (x) = f(g(x))
= f(x^1/3) =8(x³)^1/3 =8x
gof (x) = g(f(x)) = g(8x³)
= (8x³)^1/3 = 2x

Question 12.
Prove that tan^-1 \(\frac { 2 }{ 11 }\) + tan^-1 \(\frac { 7 }{ 24 }\) = tan^-1 \(\frac { 1 }{ 2 }\)
Answer:
Consider LHS : tan^-1 (\(\frac { 2 }{ 11 }\)) + tan^-1 (\(\frac { 7 }{ 24 }\))

Question 13.
Write cot^-1 \(\left(\frac{1}{\sqrt{x^{2}-1}}\right)\) ,x > 1 in the simplest form.
Answer:
Let x = sec q ⇒ q = sec^-1 x.
Now cot^-1 \(\left(\frac{1}{\sqrt{x^{2}-1}}\right)\) = cot^-1 \(\left(\frac{1}{\sqrt{\sec ^{2} \theta-1}}\right)\)
= cot^-1 \(\left(\frac{1}{\tan \theta}\right)\) = cot^-1(cot θ) = θ = sec^-1 x.

Question 14.
Find the area of the triangle with vertices (3,8), (-4, 2) and (5, 1) using determinants.
Answer:
Area of triangle

= \(\frac { 1 }{ 2 }\) [3(2 – 1) -8(-4 -5) + 1(-4 -10)]
= \(\frac { 1 }{ 2 }\) [3 + 72 -14] = \(\frac { 61 }{ 2 }\)sq.units

Question 15.
Find \(\frac{d y}{d x}\) if x² + xy + y² = 100.
Answer:
x² + xy + y² = 100 (diff wrt x)
2x + x.\(\frac{d y}{d x}\) + y.1 + 2y\(\frac{d y}{d x}\) = 0
x.\(\frac{d y}{d x}\) + 2y\(\frac{d y}{d x}\) = 2x – y
\(\frac{d y}{d x}\)(x + 2y) = -(2x + y)
\(\therefore \frac{d y}{d x}=\frac{-(2 x+y)}{x+2 y}\)

Question 16.
Find \(\frac{d y}{d x}\) if y = (logx)^cosx.
Answer:
Let y = (log x)^{cos x}
⇒ log y = cos x log (log x)
⇒ \(\frac{d }{d x}\)logy = \(\frac{d }{d x}\)cosx log(logx)

Question 17.
Find the interval in which the function f given f(x) = 2x² – 3x is strictly increasing.
Answer:
f(x) = 2x² – 3x
diff wrt x
f¹ (x) = 4x – 3
f¹ (x) = 0 ⇒ 4x – 3 = 0
or 4x = 3
⇒ x = 3/4
∴ Point x = 3/4 divides the real line into two disjoint intervals (- ∞, 3/4) and (3/4, ∞) In the interval (3/4, ∞), f¹(x) = 4x – 3 > 0
∴ f is strictly increasing in (3/4, ∞)

Question 18.
Find \(\int \frac{\left(x^{4}-x\right)^{\frac{1}{4}}}{x^{5}} d x\)
Answer:

Question 19.
Integrate x sec² x with respect to x.
Answer:

\(\int x \sec ^{2} x d x=x \int \sec ^{2} x d x-\int\left(\int \sec ^{2} x\right) \cdot \frac{d}{d x}(x) \cdot d x\)
= x. tan x – \(\int \tan x \cdot 1 \cdot d x\)
= x . tanx – log |sec x| + c

Question 20.
Find the order and degree (if defined) of the differential equation :
y¹¹¹ – y² + e³ = 0
Answer:
Order = 3, Degree = not defined.

Question 21.
If \(\overrightarrow{\mathrm{a}}\) is a unit vector and \((\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})\) = 8, then find \(|\overrightarrow{\mathrm{x}}|\) .
Answer:

Question 22.
Find the area of the parallelogram whose adjacent sides are given by the vectors \(\overrightarrow{\mathrm{a}}\) = 3î + ĵ + 4 k̂ and \(\overrightarrow{\mathrm{b}}\) = î – ĵ + k̂
Answer:
Area of parallelogram = \(|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|\)

Question 23.
Find the angle between the pair of lines \(\frac{x+3}{3}=\frac{y-1}{5}=\frac{z+3}{4}\) and \(\frac{x+1}{1}=\frac{y-1}{1}=\frac{z-5}{2}\)
Answer:
Given
\(\frac{x+3}{3}=\frac{y-1}{5}=\frac{z+3}{4}\)
Here \(\overrightarrow{\mathrm{b}}_{1}\) = 3î + 5ĵ + 4k̂
and \(\frac{x+1}{1}=\frac{y-1}{1}=\frac{z-5}{2}\) ⇒ î + ĵ + 2k̂

Question 24.
The probability distribution of random variable X is as follows :

Answer:
Here E(X) = Σx_i P(x_i)
\(=0 \times \frac{188}{221}+1 \times \frac{32}{221}+2 \times \frac{1}{221}=\frac{32+2}{221}=\frac{34}{221}\)

Part-C

Answer any TEN questions: (10 × 3 = 30)

Question 25.
Determine whether the relation R in the set A = {1, 2, 3, …., 13, 14} defined as R = {(x,y); 3x – y = 0} is reflexive, symmetric and transitive.
Answer:
Given A = {1,2,3, ….,13, 14}
R = {(x, y) : 3x – y =0}
∴ R= {1,3), (2, 6), (3, 9), (4, 12)}
Here (1, 1) ∉ R, (2, 2) ∉ R
R is not reflexive
Here (1, 3) ∈ R but (3, 1) ∉ R
∴ R is not symmetric
Here (1,3) and (3, 9)6 ∈ R but (1,9) ∉ R
∴ R is not transitive
∴ R is not an equivalence relation.

Question 26.
Solve for x, if tan^-1 2x + tan^-13x = \(\frac{\pi}{4}\), x > 0
Answer:
Given tan^-1(2x) + tan^-1(3x) = \(\frac{\pi}{4}\)
tan^-1 \(\left(\frac{2 x+3 x}{1-2 x \cdot 3 x}\right)\) = \(\frac{\pi}{4}\)
\(\frac{5 x}{1-6 x^{2}}\) = tan \(\frac{\pi}{4}\)
⇒ \(\frac{5 x}{1-6 x^{2}}\) = 1 ⇒ 5x = 1 – 6x²
∴ 6x² + 5x – 1 = 0
6x² + 6x – x – 1 =0
6x(x+1) -1(x+1) = 0 or (x+1) (6x – 1)= 0
⇒ x = -1 or x = \(\frac { 1 }{ 6 }\)

Question 27.
By using elementary transformations, find the inverse of the matrix

Answer:

Question 28.
If x = a(θ + sinθ), y = a( 1 – cosθ) then show that \(\frac{d y}{d x}\) = tan \(\left(\frac{\theta}{2}\right)\).
Answer:
we have
\(\frac{d x}{d \theta}\) = a(1 + cosθ), \(\frac{d y}{d \theta}\) = a(sin θ)

Question 29.
Verify mean value theorem for the function f(x) = x² – 4x – 3 in the interval [1, 4].
Answer:
f(x) = x² – 4x – 3 interval [a, b] = [1,4]
Here f is a polynomial function hence continuous in [1, 4] and differentiable in (1,4)
f(x) = x² – 4x – 3 (diff wrt x)
∴ f¹(x) = 2x – 4 1
Also f(1) = 1² – 4 – 3 = -6
f(4) = 4² – 4(4) – 3 = -3
∴ From MVT ∃ c ∈ (a,b) such that
f¹ (c) = \(\frac{f(b)-f(a)}{b-a}=\frac{-3-(-6)}{4-1}\)
f¹ (c) = \(\frac { 3 }{ 3 }\) = 1
f¹ (c) = 2c – 4
∴ 2c – 4 = 1
or 2c = 5
c = \(\frac { 5 }{ 2 }\) ∈ (1, 4)

Question 30.
Find the point, at which the tangent to the curve y = \(\sqrt{4 x-3}-1\) has its slope \(\frac { 2 }{ 3 }\)
Answer:
y = \(\sqrt{4 x-3}-1\) (diff wrt x)
\(\frac{d y}{d x}=\frac{1}{2 \sqrt{4 x-3}} \cdot 4\)
Given slope of tangent = \(\frac{d y}{d x}=\frac{2}{3}\)
∴ \(\frac{2}{3} =\frac{1}{2 \sqrt{4 x-3}} .4\)
\(\frac{4}{3} =\frac{4}{\sqrt{4 x-3}} \Rightarrow \sqrt{4 x-3}=3\)
\(\sqrt{4 x-3} = 3\) (squaring)
4x – 3 = 9
4x = 12
0r x = 3
∴ y = \(\sqrt{4(3)-3-1}\)
= \(\sqrt{9}\) – 1 = 3 – 1 = 2
∴ point = (3, 2)

Question 31.
Find \(\int \frac{x}{(x+1)(x+2)} d x\).
Answer:

Question 32.
Evaluate \(\int_{0}^{2}\left(x^{2}+1\right) d x\). as a limit of a sum.
Answer:
\(\int_{0}^{2}\left(x^{2}+1\right) d x\) Here a = 0 ; b = 2
f (x) = x² + 1

Question 33.
Find the area of the region bounded by y² = 9x, x = 2, x = 4 and the x – axis in the first quadrant.
Answer:

Question 34.
If y = ae^3x + be^-2x represents family of curves, where a and b are arbitrary constants. Form the differential equation.
Answer:
Given, y = ae^3x + be^-2x …. (1)
y’ = (ae^3x)3 – (be^-2x) 2
y’ = 3a e^3x – 2be^-2x …. (2)
From (1) and (2)
y = ae^3x + be^-2x × 3
y’ = 3a e^3x – 2b e^-2x × 1

Diff
⇒ 3y’ – y” = 5be^-2x
3y’ – y”= -10be^-2x …. (4)
From (3) & (4)
3y’ – y” = (-2) (5be^-2x)
3y’ – y” = -2[3y – y’]
3y’- y” = -6y + 2y’
y” – y’ – 6y= 0.

Question 35.
Show that the position vector of the point P, which divides the line joining the points A and B having position vectors \(\overrightarrow{\mathrm{a}}\) and \(\overrightarrow{\mathrm{b}}\) internally in the ratio
\(\mathbf{m}: \mathbf{n} \text { is } \frac{\mathbf{m} \overrightarrow{\mathbf{b}}+\mathbf{n} \overrightarrow{\mathbf{a}}}{\mathbf{m}+\mathbf{n}}\)
Answer:
Here P divides AB internally in the ratio m : n

∴ p divides \(\overrightarrow{\mathrm{AB}}\) such that
\(\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{m}}{\mathrm{n}}\) or n\(\overrightarrow{\mathrm{AP}}\) = m\(\overrightarrow{\mathrm{PB}}\)
Where m and n are positive scalars
Here \(\overrightarrow{\mathrm{r}}, \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}\) are position vectors of \(\overrightarrow{\mathrm{OP}}\)
\(\overrightarrow{\mathrm{OA}}\) an \(\overrightarrow{\mathrm{OB}}\) respectively
∴ From the figure we have,

Question 36.
Prove that \([\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}, \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{a}}]\) = 2\([\vec{a}, \vec{b}, \vec{c}]\)
Answer:
LHS

Question 37.
Find the shortest distance between the lines \(\overrightarrow{\mathbf{r}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\lambda(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})\) \(\overrightarrow{\mathbf{r}}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}+\mu(3 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})\)
Answer:
2nd PUC Maths Previous Year Question Paper June 2019 27

Question 38.
Bag I contains 3 red and 4 black balls. White Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find
Answer:
Let A be the event of choosing the bag I
B be the event of choosing the bag II
and C be the event of drawing a red ball
∴ P(A) = P(B) = \(\frac { 1 }{ 2 }\)
Also p(C/A) = \(\frac { 3 }{ 7 }\) and P(C/B) = \(\frac { 5 }{ 11 }\)
∴ P(C)=P(A).P(C/A) +P(B).P(C/B)
\(=\frac{1}{2} \times \frac{3}{7}+\frac{1}{2} \times \frac{5}{11}=\frac{3}{14}+\frac{5}{22}=\frac{33+35}{154}\)
P(C) = \(\frac { 68 }{ 154 }\)
By Baye’s Theorem,

P(B/C) = \(\frac { 35 }{ 68 }\)

Part-D

Answer any SIX questions: (6 × 5 = 30)

Question 39.
Prove that the function f = R → R defined by f(x) = 4x + 3, is invertible and find the inverse of f.
Answer:
Let f(x₁) = f(x₂)= 4x₁ + 3 ⇒ 4x₂ + 3
⇒ 4x₁ = 4x₂ ⇒ x₁ = x₁
f is onto.
Let y ∈ R∃ x ∈ R such that f(x) = y
4x + 3 = y \(x=\frac{y-3}{4} \in R\)
∴ f is onto ∴ f is bijection
∴ f is invertible
Define a function g : R → R by
g(y) = \(\frac{y-3}{4}\)
(fog) (y) = f(g(y)) = f\(\frac{y-3}{4}\)
= 4\(\frac{y-3}{4}\)+3 = y – 3 + 3 = y ∴ (fog) (y) = y fog = I
(gof) (x) = g(f(x)) = g (4x + 3)
\(=\frac{4 x+3-3}{4}=\frac{4 x}{4}=x\)
∴ (gof) (x) = x ∴gof = I
∴ f^-1(y) = g(y) = \(\frac{y-3}{4}\) ∀ y ∈ R

Question 40.
If


Calculate AC, BC and (A + B)C.
Answer:

Question 41.
Solve the following system of linear equations by matrix method:
3x – 2y + 3z = 8
2x + y – z = 1
4x – 3y + 2z = 4
Answer:
The system of equations can be written in the form A X = B where

|A| = 3 (2 – 3) + 2(4 + 4) + 3 (-6 -4) = – 17
Hence, A is non singular and so its inverse exists. Now
A₁₁ = -1, A₁₂ = -8 A₁₃ = -10
A₂₁ = -5, A₂₂ = -6 A₂₃ = +1
A₃₁ = -1, A₃₂ = +9 A₃₃ = +7
A^-1 = \(\frac{1}{|\mathrm{A}|}\) adj A

substituting in (1)

Question 42.
If y = 3 cos (log_ex) + 4 sin (log_ex) show that x²y₂ + xy₁ + y = 0.
Answer:
y = 3 cos (log_ex) + 4sin (log_ex)
Diff wrt x
y₁ = 3(-sin(logx).\(\frac { 1 }{ x }\)) + 4 cos (log x). \(\frac { 1 }{ 2 }\)
xy₁ = -3 sin (logx) + 4 cos(logx)
diff wrt x
xy₂ y₁. 1 = -3cos(logx).\(\frac { 1 }{ x }\) + 4(-sin(logx)).\(\frac { 1 }{ x }\)
xy₂ + y₁ = \(\frac { 1 }{ 2 }\)-(3cos(logx) + 4sin(logx))
x²y₂ + xy₁ = -(3cos(logx) + 4sin(logx))
x²y₂ + xy₁ = -y
∴ x²y₂+ xyj₁+ y = 0

Question 43.
Sand is pouring from a pipe at the rate of 12cm3/sec. The falling sand forms a cone on the ground in such a way that the height of the cone is always one – sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?
Answer:
Volume, V = \(\frac{\pi r^{2} h}{3}\)
given; h = \(\frac { 1 }{ 6 }\) r (or) r = 6h
∴ v = \(\frac{\pi}{3}\) × (6h)² ×h = 12πh³
v = 12πh³
diff w.r.t ‘t’

Question 44.
Find the integral of \(\sqrt{x^{2}+a^{2}}\) with respect to x, and hence find \(\int \sqrt{x^{2}+2 x+5} d\)
Answer:

Question 45.
Using integration find the area of the region in the first quadrant enclosed by the x – axis, the line y = x, ad circle x² + y² = 32
Answer:

x² + y² = 32 and centre (0,0)
∴ radius = r = \(\sqrt{32}\) = \(4 \sqrt{2}\)
Equation are y = x ….(1) and x² + y² = 32 …..(2)
∴x² + x² = 32 (usinig (1) and (2))
∴ 2x² = 32 ⇒ x² = 16 or x = ±
since region is in first quadrant
x = 4 ⇒ y = 4
∴ Required Area = Area OBM + Area MBA

Question 46.
Find the general solution of the differential equation (x+ y)\(\frac{d y}{d x}\) = 1
Answer:
(x + y)\(\frac{d y}{d x}\) = l dx
(x + y)dy = dx or \(\frac{d y}{d x}\) = x + y
∴ \(\frac{d y}{d x}\) -x = y which i s in the form
\(\frac{d y}{d x}\) + Px =Q dy
Here P = -1; Q = y
∴ Integrating Factor = IF = \(e^{\int P d y}\)
\(=e^{\int(-1) d y}=e^{-y}\)
∴ General solution is
x(IF) = \(\int y \cdot(I F) d y+C\) (Using integration by parts)
= y(-e^-y) – \(\int-\left(e^{-y}\right)\).1.dy + C
= -ye^-y +\(\int e^{-y} d y\) + C ÷ e^-y
xe^-y = -ye^-y – e^-y + C
⇒ x = -y-l+C.e^y
Or x + y + 1 = C.e^y

Question 47.
Derive the equation of a plane in normal form both in the vector and Cartesian form.
Answer:

Consider a plane whose perpendicular distance from the origin is d. where d ≠ 0.


This is the Required Equation of a plane m normal form.
To find P and Q :
x² = 4y and x =4y – 2
=>(4y – 2)² = 4y ⇒ 16y² + 4 – 16 y=4 y
⇒ 16y² – 20y +4 = 0
⇒ (y – 1) (16y – 4) = 0
⇒ y = 1 (or) y = \(\frac { 1 }{ 4 }\)
if y = 1 ; x = 2 ⇒ 2 (2,1)
and y = \(\frac { 1 }{ 4 }\) ; x = -1 ⇒ (-1, \(\frac { 1 }{ 4 }\))
Thus the points of intersection are
(2,1) I(-1 .\(\frac { 1 }{ 4 }\))
∴ Area of the Region O B Q C = Area of BCQ + Area of QCO

Question 48.
A die is thrown 6 times. If ‘getting an odd number’ is success, what is the probability of
a) 5 successes
b) at least 5 successes
c) at most 5 Successes
Answer:
Let P = probability of getting an odd number = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)
q = 1 – p = 1\(\frac { -1 }{ 2 }\) = \(\frac { 1 }{ 2 }\)
& n =6
we have P(x) = ⁿc_x P^x q^n-x

Part-E

Answer any ONE question: (1 × 10 = 10)

Question 49.
a) Prove that \(\int_{0}^{b} f(x) \cdot d x=\int_{a}^{b} f(a+b-x) d x\) and hence. Evaluate \(\int_{\pi / 3}^{\pi / 3} \frac{1}{1+\sqrt{\tan x}} d x\) (June – 2014,Q.No.49a)
Solution :

let t = a + b – x then dt = -dx
when x =a and x = b
t = b t = a

f(x) is continuous, find the value of k.

Question 50.
(a) A factory manufactures two types of screws, A and B. Each type of Screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at most 4 hours on any day. The manufacture can sell a package of screws A at a profit of Rs.7 and seres B at a profit of Rs.10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximise his profit? Determine the maximum profit.
Answer:

∴ Maximise Z = 7x + 10y subject to constraints
4x + 6y ≤ 240
or 2x + 3y ≤ 120 …… (1)
and 6x + 3y ≤ 240
or 2x + y ≤ 80 …… (2)
and x, y ≥ 0
Converting inequalities to equalities
2x + 3y= 120

2x + y = 80

Plot the graph


Maximum profit is Rs.410 which occurs at point B (30, 20)
∴Manufacturer should manufacture 30 packages of screws A and 20 packages of screws B.

Answer:


= (1 + x + x²)(1 – x²) [1(1 + x + x²) – 0 + 0] (Expanding along c₁)
= (1 + x + x²) (1 + x + x²) (1 – x²)
= [(1 + x + x²) (1 – x²)]² = (1³ – x³)² (using a³ – b³
= (a – b) (a² + b² +ab))
= (1 – x³)² (RHS)